2  

I highly recommend to use this regex which will capture what you want:

我强烈建议使用这个正则表达式来捕获你想要的东西：

re.findall(r'parsable-cite=\\\"(.*?)\\\"\>',page)

explanation:

说明：

parsable-cite= matches the characters parsable-cite= literally (case sensitive)
  \\ matches the character \ literally
  \" matches the character " literally
  1st Capturing group (.*?)
  .*? matches any character (except newline)
      Quantifier: Between zero and unlimited times, as few times as possible,
           expanding as needed
  \\ matches the character \ literally
  \" matches the character " literally
  \> matches the character > literally

using ? is the key ;)

使用？是关键;）

hope this helps.

希望这可以帮助。

1  

Make your regex lazy:

让你的正则表达式懒惰：

re.findall(r'^parsable-cite=.*?>$',page)
                              ^

Or use a negated class (preferable):

或者使用否定的类（最好）：

re.findall(r'^parsable-cite=[^>]*>$',page)

.* is greedy by default and will try to match as much as possible before concluding a match.

。*默认情况下是贪婪的，并会在结束比赛前尽可能地匹配。

regex101 demo

regex101演示

If you want to get the parts you need only, you can use capture groups:

如果您只想获得所需的零件，可以使用捕获组：

re.findall(r'^parsable-cite=([^>]*)>$',page)

regex101 demo

regex101演示

Though, from the layout of your webpage, it doesn't seem that you need the anchors (^ and $) (unless the newlines were somehow removed on the site...)

但是，从您的网页布局来看，您似乎不需要锚点（^和$）（除非在网站上以某种方式删除换行符...）

1  

The .* you have there is "greedy", meaning it will match as much as it can, including any number of > characters and whatever comes after them.

。*你有“贪婪”，这意味着它将尽可能多地匹配，包括任意数量的>字符以及它们之后的任何内容。

If what you really want is "everything up to the next >" then you should say [^>]*> instead, meaning "any number of non-> characters, then a >".

如果你真正想要的是“一切都是下一个>”那么你应该说[^>] *>，意思是“任意数量的非>字符，然后是>”。

1  

maybe something like this:

也许是这样的：

(?<=parsable-cite=\\\")\w{2}\/\d{3}\/\d{3}

http://regex101.com/r/kE9uE3

http://regex101.com/r/kE9uE3

1  

Though this is a json string where html is embedded inside, but you can still use BeautifulSoup for this purpose:

虽然这是一个json字符串，里面嵌入了html，但你仍然可以使用BeautifulSoup来达到这个目的：

soup = BeautifulSoup(htmls);
tags = soup.findAll("external-xref", {"parsable-cite":re.compile("")})
for t in tags:
    print t['parsable-cite']

1  

This might work if its between \" delimiters

如果它在“分隔符之间”，这可能会有效

 #  \bparsable-cite\s*=\s*\"((?s:(?!\").)*)\"

 \b 
 parsable-cite
 \s* = \s* 
 \"
 (                             # (1 start)
      (?s:
           (?! \" )
           . 
      )*
 )                             # (1 end)
 \"

Or, just

要不就

 #  (?s)\bparsable-cite\s*=\s*\"(.*?)\"

 (?s)
 \b 
 parsable-cite
 \s* = \s* 
 \"
 ( .*? )                 # (1)
 \"

1  

If you think it will be very similar each time:

如果您认为每次都非常相似：

re.findall(r"pl/\d+/\d+", page)